Problem: $f(x)=\dfrac{18}{\sqrt[{3}]{x-2}}$. On which intervals is the graph of $f$ concave down? Choose 1 answer: Choose 1 answer: (Choice A) A $x>\sqrt[3]{2}$ only (Choice B) B $x>-2$ only (Choice C) C $x<2$ only (Choice D) D $x<-\sqrt[3]{2}$ only
We can analyze the intervals where $f$ is concave up/down by looking for the intervals where its second derivative $f''$ is positive/negative. This analysis is very similar to finding increasing/decreasing intervals, only instead of analyzing $f'$, we are analyzing $f''$. The second derivative of $f$ is $f''(x)=\dfrac{8}{\sqrt[3]{(x-2)^7}}$. $f''$ is never equal to $0$. $f''$ is undefined for $x=2$. Therefore, our only point of interest is $x=2$. Our point of interest divides the domain of $f$ (which is all numbers except for $2$ ) into two intervals: $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $x<2$ $x>2$ Let's evaluate $f''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f''(x)$ Verdict $x<2$ $x=1$ $f''(1)=-8<0$ $f$ is concave down $\cap$ $x>2$ $x=3$ $f''(3)=8>0$ $f$ is concave up $\cup$ In conclusion, the graph of $f$ is concave down over the interval $x<2$ only.